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Math Help - ddifferentiation using l'Hopitals rule

  1. #1
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    differentiation using l'Hopitals rule

    Hello, I am trying to diffferentiate using l'Hopital's rule:
    (e^(5+h)-e^5) / (h)
    to find the limit of h tending towards zero.

    I differentiate the numerator and denominator to get:

    (e^(5+h) - e^5) / (1)

    but I think I should be getting (e^(5+h)) / (1)
    Can someone please tell me how the second e^5 is done away with ?

    Thanks kindly.
    Last edited by fran1942; April 3rd 2012 at 05:39 PM.
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  2. #2
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    Re: ddifferentiation using l'Hopitals rule

    You don't differentiate using L'Hospital's Rule, you differentiate the top and the bottom because that's what you need to do to USE L'Hospital's Rule. Anyway

    \displaystyle \begin{align*} \lim_{h \to 0}\frac{e^{5 + h} - e^5}{h} &= \lim_{h \to 0}\frac{e^5e^h - e^5}{h} \\ &= \lim_{h \to 0}\frac{e^5\left(e^h - 1\right)}{h} \\ &= e^5\lim_{h \to 0}\frac{e^h - 1}{h} \textrm{ which goes to }\frac{0}{0}\textrm{ so L'Hospital's Rule can be used} \\ &= e^5\lim_{h \to 0}\frac{\frac{d}{dx}\left(e^h - 1\right)}{\frac{d}{dx}\left(h\right)} \textrm{ by L'Hospital's Rule} \\ &= e^5\lim_{h \to 0}\frac{e^h}{1} \\ &= e^5\lim_{h \to 0}e^h \\ &= e^5 \cdot 1 \\ &= e^5 \end{align*}
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  3. #3
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    Re: differentiation using l'Hopitals rule

    Quote Originally Posted by fran1942 View Post
    Hello, I am trying to diffferentiate using l'Hopital's rule:
    (e^(5+h)-e^5) / (h)
    to find the limit of h tending towards zero.

    I differentiate the numerator and denominator to get:

    (e^(5+h) - e^5) / (1)

    but I think I should be getting (e^(5+h)) / (1)
    Can someone please tell me how the second e^5 is done away with ?

    Thanks kindly.

    Circular reasoning
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  4. #4
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    Re: ddifferentiation using l'Hopitals rule

    e^5 is a constant and differentiates to zero
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